Kummer's transformation of series

Template:Short description In mathematics, specifically in the field of numerical analysis, Kummer's transformation of series is a method used to accelerate the convergence of an infinite series. The method was first suggested by Ernst Kummer in 1837.

Technique

Let $$A=\sum _{n=1}^{\mathrm{\infty }}{a}_n$$ be an infinite sum whose value we wish to compute, and let $$B=\sum _{n=1}^{\mathrm{\infty }}{b}_n$$ be an infinite sum with comparable terms whose value is known. If the limit $$\gamma :=\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}$$ exists, then $a_n-\gamma b_n$ is always also a sequence going to zero and the series given by the difference, ${\sum }_{n=1}^{\mathrm{\infty }}(a_n-\gamma b_n)$, converges. If $\gamma \ne 0$, this new series differs from the original ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n$ and, under broad conditions, converges more rapidly. We may then compute $A$ as

$$A=\gamma B+\sum _{n=1}^{\mathrm{\infty }}(a_n-\gamma b_n),$$

where $\gamma B$ is a constant. Where $a_n\ne 0$, the terms can be written as the product $(1-\gamma b_n/a_n)a_n$. If $a_n\ne 0$ for all $n$, the sum is over a component-wise product of two sequences going to zero,

$$A=\gamma B+\sum _{n=1}^{\mathrm{\infty }}(1-\gamma b_n/a_n)a_n$$.

Example

Consider the Leibniz formula for π: $$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots =\frac{\pi }{4}.$$ We group terms in pairs as $$\begin{array}{cc}& 1-(\frac{1}{3}-\frac{1}{5})-(\frac{1}{7}-\frac{1}{9})+\cdots \\ & =1-2(\frac{1}{15}+\frac{1}{63}+\cdots )=1-2A\end{array}$$ where we identify $$A=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{16n^2-1}.$$ We apply Kummer's method to accelerate $A$, which will give an accelerated sum for computing $\pi =4-8A$.

Let $$\begin{array}{cc}B& =\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4n^2-1}=\frac{1}{3}+\frac{1}{15}+\cdots \\ & =\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\cdots \end{array}$$ This is a telescoping series with sum value Template:Frac. In this case $$\gamma :=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{16n^2-1}}{\frac{1}{4n^2-1}}=\underset{n\to \mathrm{\infty }}{lim}\frac{4n^2-1}{16n^2-1}=\frac{1}{4}$$ and so Kummer's transformation formula above gives $$\begin{array}{cc}A& =\frac{1}{4}\cdot \frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1}{4}\frac{\frac{1}{4n^2-1}}{\frac{1}{16n^2-1}})\frac{1}{16n^2-1}\\ & =\frac{1}{8}-\frac{3}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{16n^2-1}\frac{1}{4n^2-1}\end{array}$$ which converges much faster than the original series.

Coming back to Leibniz formula, we obtain a representation of $\pi$ that separates $3$ and involves a fastly converging sum over just the squared even numbers $(2n{)}^2$, $$\begin{array}{cc}\pi & =4-8A\\ & =3+6\cdot \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(4(2n{)}^2-1)((2n{)}^2-1)}\\ & =3+\frac{2}{15}+\frac{2}{315}+\frac{6}{5005}+\cdots \end{array}$$

See also

• Euler transform

References

• Template:SpringerEOM
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External links

• Template:MathWorld

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